This is complete garbage. The first equation, v = Em, is not even dimensionally correct. No offense, but whoever wrote this doesn't have a clue what he/she is talking about. If you want a derivation for E=mc2, consult just about any book on special relativity. If you would like me to post one, let me know.

Regarding p = E/c for light. This can be deduced in a few ways. One of which is from E2 - (pc)2 = m02c4. Use this for a single photon and the fact that the proper mass, m0, is zero. That yields p = E/c. This relationship can be derived from the principles of electromagnetism by requiring both energy and momentum be conserved. I think I did this out on my website but geocities doesn't seem to be working now.

2) Surely the distance travelled by the light is NOT equal to L, since the object is recoiling in the other direction so this distance will be shorter?

The distance is so small that its ignored.

3) And thirdly, why is Mx= mL?

It represents the conservation of the center of mass of the system which must remain fixed. In fact Einstein used E = mc2 to show that the center of mass remains fixed in his paper The Principle of the Conservation of Motion of the Center of Gravity and the Inertia of Energy, A. Einstein, Annalen der Physik 20 (1906): 627-233.

Note: Einstein mentions the work of Poincare in that paper. Poincare spoke of the mass of light as does the link you provided. Einstein writes

Although the simple formal considerations that have to be carried out to prove this statement are in the main already contained in a work by H. Poincare, for the sake of clarity I shall not base myself upon that work.

Centre of mass location of 2 masses m_1 and m_2 is (m_1x_1+m_2x_2)/(m_1+m_2). For this not to change requires[tex]m_1\Delta x_1=-m_2\Delta x_2[/tex]. Get it now?

BTW, this is not a proof, but only demonstrates consistency.

BTW, it has a glaring hole, as it relies on an infinitely rigid cylinder. IOW, when the light is emitted, the whole cylinder cannot instantaneously begin to move, in fact the other end only "feels" the movement after a delay equal to the time of transit of a sound wave through the cylinder material; this is very much longer time than that needed for the light pulse to traverse it. Improved versions of this gedanken experiment are available in better text books.

L could be the distance traveled in recoil, theoretically, IF the energy conveyed that caused the recoil were purely directed along the reverse vector to the direction of primary projection and IF no energy were wasted or transferred along any other axis - and of course assuming away things like friction, resistance, and shifts in gravity as the energy travels.

I'm beginning to wonder about c (light speed). The E-Mc^2 is predicated on light speed being constant. NO longer sure abou that. From our perspective, sitting in our gravity well (Earth), sure, looks constant, even with the tools we have that measure to the nth decimal point. But what if you measured light speed from the surface of Mercury (if you weren't melted by being that close to the sun and stuff)? Far closer to a star and on a smaller planetary mass, would light seem to be traveling at the exact same speed.

I'm beginning to wonder about c (light speed). The E-Mc^2 is predicated on light speed being constant. NO longer sure abou that. From our perspective, sitting in our gravity well (Earth), sure, looks constant, even with the tools we have that measure to the nth decimal point. But what if you measured light speed from the surface of Mercury (if you weren't melted by being that close to the sun and stuff)? Far closer to a star and on a smaller planetary mass, would light seem to be traveling at the exact same speed.

Hey NASA!

Short answer: Yes.

Longer answer: Sitting on the surface on a neutron star, free falling towards a black hole, floating about in space, inside an accelerating rocket, inside a high speed centrifuge, the speed of light in a vacuum is always c if measured locally in a small enough region.

So, that being said, doesn't that mean that E=Mc^2 can't be right, because it's only RELATIVELY correct? If our tech is moved away from our location relative to the Earth and Sun etc., shouldn't it be made to deal with the shift in gravatic consequence? I just mean: are we building spacecraft that go to Jupiter and beyond with an inherant flaw due to simply trusting E=Mc^2?

This is complete garbage. The first equation, v = Em, is not even dimensionally correct. No offense, but whoever wrote this doesn't have a clue what he/she is talking about. If you want a derivation for E=mc2, consult just about any book on special relativity. If you would like me to post one, let me know.

- Warren

You are right v=Em is wrong and it cant be derived the true derevation is that v=E/Mc.